Let $X\subseteq\mathbb{R}$ be an open subset and $E$ be a Banach space.
Denote $C_0^1(X,E)$ as the space of $E$-valued functions $f$ on $X$ such that $f,f'$ are continuous, and $f$ vanish at infinity. Also, denote $\text{supp }f=\overline{\{ x\in\mathbb{R}:f(x)\neq 0 \}}$
Question: Fix two linear functionals $y_0^*$ and $y_1^*$ on $E$ with $$\| y_0^* \|^2 +\| y_1^* \|^2 = 1.$$ Given any $\varepsilon>0,\delta>0$ and $x\in X,$ does there exist a function $f\in C_0^1(X,E)$ such that $$\sup_{z\in X}\sqrt{(\|f(z)\|^2+\|f'(z)\|^2)}=1, $$ $$\text{ supp} f \subseteq (x-\delta,x+\delta)$$ and $$y_0^*(f(x)) + y_1^*(f'(x))>1-\varepsilon?$$
Let $y^*:=(y_0^*, y_1^*)$ be a function $y^*:E^2_{\ell^2}\to\mathbb{R}.$ By assumption, $\|y^*\|_{\ell^2}=1.$
Choose $y:=(y_0,y_1)\in E^2_{\ell^2}$ with $\|y\|_{\ell^2}=1$ such that $$y_0^*(y_0) + y_1^*(y_1)>1-\varepsilon.$$
Now, it remains to construct a function $f\in C_0^1(X,E)$ such that its norm is $1,$ supported in $(x-\delta,x+\delta)$ and $f(x)=y_0,f'(x)=y_1.$
However, I have no idea on how to construct such function.
The comment by Daniel Fischer already gives the answer. A more precise statement is: If you fix $\delta>0$ then such a function exists for any $\epsilon>0$ only if $\|y^*_0\|\leq \sin (\delta)$. In particular, this can only hold for any $\delta>0$ if $y^*_0=0$.
For simplicity of notation assume $x=0$.
The condition: $|f(z)|^2+|f'(z)|^2\leq 1$ implies: $|f'(z)|\leq \sqrt{1-|f(z)|^2}$. Since $f(\pm \delta)=0$ it follows first that $|f(z)|\leq \sin (\delta-|z|)$, $|z|\leq \delta$ and then that $$ |y_0^* (f(z)) + y_1^* (f'(z)) | \leq |y_0^*| \sin (\delta-|z|) + |y_1^*| \cos(\delta-|z|) $$ When $\|y_0^*\|>\sin(\delta)$ the RHS is uniformly bounded below 1, thus preventing the construction of a solution.
On the other hand, when $y_0^*=0$ you may construct solutions by picking a unit vector $y_1$ with $y_1^*(y_1)=1$ and a $C^1$ function $\phi:{\Bbb R}\rightarrow [0,1] $ with support in $(-1,1)$ and $\phi'(0)=\sup|\phi'|=1$. Construct for $0<t<\delta$: $$f_t(z) = y_1 \frac{t}{\sqrt{1+t^2}} \phi(z/t)$$ One has $\sup(\|f_t\|^2 + \|f'_t\|^2)=1$ and as $t$ goes to zero, $\sup_z\|f_t(z)\|\rightarrow 0$ and $f_t'(0)\rightarrow y_1$. Given $\epsilon>0$ the result follows for $t$ small enough.